can't understand why integral of xe^((x^2)/2) is e^((x^2)/2)
can someone explain how to calculate this integral please?

- anonymous

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- UsukiDoll

\[\LARGE xe^{\frac{x^2}{2}}\] this one?

- UsukiDoll

do you know u-substitution?

- UsukiDoll

The u-substitution is an integration method that essentially involves using the chain rule in reverse.

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## More answers

- UsukiDoll

hmm let's try rewriting that fraction
\[\LARGE xe^{\frac{1}{2}x^2}\] so we can see the exponent clearly.

- UsukiDoll

we need to pick a u so that when we take that derivative of u we can substitute..
so what will be our u?
Well we do see an x in front of the e.. Maybe we can have
\[\LARGE u=\frac{1}{2}x^2 \]

- UsukiDoll

so what is the derivative of \[\large \frac{1}{2}x^2 ? \]

- UsukiDoll

we can use product rule to make this a bit easier. Let f(x) = 1/2 and g(x) = x^2
recall that the product rule is f(x)g'(x) +f'(x)g(x)

- UsukiDoll

so what is the derivative of f(x) = 1/2 ?

- UsukiDoll

similarly what is the derivative of \[\large g(x) = x^2 \]

- anonymous

so if \[u= (1/2)x^2 -> u' = x\]

- UsukiDoll

yeah but we have to write this differently
it will be
du = x dx

- anonymous

so now we have \[xe^u dx\]

- UsukiDoll

hmm how do I explain this.... we just have \[e^u \]
and substitute the u back .. don't forget to add +C

- UsukiDoll

remember du = x dx ?
since there's nothing on the left hand side.... oh man .. I'm having trouble explaining this part. . . usually when there's nothing ... it's just blank like 1 sorry I'm in panic mode I can't figure out how to type this well

- UsukiDoll

but we just have \[\LARGE e^u +C \] now substitute \[\LARGE u=\frac{1}{2}x^2 \]
back into
\[\LARGE e^u +C \]

- anonymous

ok, so \[xe^u dx -> xe^u (1/x) dy -> e^u du\] (+c...)

- anonymous

du

- UsukiDoll

you don't have to write dx
you don't have to write du

- UsukiDoll

du = x dx
it's just .... @ganeshie8 can you help me explain this?
I know what's going on just can't put it into words

- anonymous

Ok, I think I understand, thank you very much!

- UsukiDoll

but in the end you should have \[\LARGE e^u +C \] now plug in u back into this equation

- UsukiDoll

and once we have our antiderivative, we can simply check by taking the derivative.

- UsukiDoll

you should have something like \[\LARGE e^{\frac{1}{2}x^2} +C\]

- UsukiDoll

now take the derivative of that equation...

- UsukiDoll

that's how to check :)

- anonymous

so we get back to \[e^((x^2)/2) \]

- anonymous

yes, +c

- UsukiDoll

mhm

- UsukiDoll

if you take the derivative of your antiderivative, we should be getting that x in front of that e

- anonymous

Yes, I wrote the integral

- UsukiDoll

\[\LARGE f(x) = e^{\frac{1}{2}x^2} +C \]
\[\LARGE f'(x) = xe^{\frac{1}{2}x^2} +C \]
because the derivative is \[\frac{1}{2}x^2 \rightarrow \frac{1}{2} (2x) + 0(x^2) \rightarrow x\] and that's placed in front of the e.

- UsukiDoll

the derivative of any e^x is always in the front and the exponent part of e^x just stays there :)

- anonymous

Excellent. Thanks a lot!

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